Question

Выполните умножение дробей а)9с^3/а*а^2/12с^3 б)а^4с/8b*12b^3/a^3c^4 в)6x*b^5/3b3 выполните действия а)(3/x-3-x+15/x^2-9): x/x+3 b)(4/y-2/y-5): y/y^2-25+5/y+5 !

Answer 1

А)$\frac{9c^3}{a}* \frac{a^2}{12c^3}= \frac{9c^3a^2}{12ac^3}= \frac{3a}{4}$
б)$\frac{a^4c}{8b}* \frac{12b^3}{a^3c^4} = \frac{12a^4b^3c}{8a^3bc^4} = \frac{3ab^2}{2c^3}$
в)$\frac{6xb^5}{3b^3}=2xb^2$
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а)$( \frac{3}{x-3} - \frac{x+15}{x^2-9}): \frac{x}{x+3} = \frac{3(x+3)-x-15}{(x-3)(x+3)}: \frac{x}{x+3}= \frac{(2x-6)(x+3)}{(x-3)(x+3)x}= \frac{2(x-3)}{x(x-3)}= \frac{2}{x}$
b)$(\frac{4}{y}- \frac{2}{y-5}): \frac{y}{y^2-25}+ \frac{5}{y+5}= \frac{4y-20-2y}{y(y-5)}: \frac{y}{(y-5)(y+5)}+ \frac{5}{y+5}$$= \frac{(2y-20)(y-5)(y+5)}{y^2(y-5)}+ \frac{5}{y+5}= \frac{(2y-20)(y+5)}{y^2}+ \frac{5}{y+5}= \frac{(2y-20)(y+5)^2+5y^2}{y^2(y+5)}=$$\frac{(2y-20)(y+5)^2+5y^2}{y^2(y+5)}= \frac{2y^3+20y^2+50y-20y^2-200y-500}{y^2(y+5)}= \frac{2y^3-150y-500}{y^2(y+5)}$